Google
Câu hỏi: Cho $a,b>0$, nếu ${{\log }_{8}}a+{{\log }_{4}}{{b}^{2}}=5$ và ${{\log }_{4}}{{a}^{2}}+{{\log }_{8}}b=7$ thì giá trị của $ab$ bằng:
A. ${{2}^{9}}$.
B. $2$.
C. $8$.
D. ${{2}^{18}}$.
A. ${{2}^{9}}$.
B. $2$.
C. $8$.
D. ${{2}^{18}}$.
Ta có: $\left\{ \begin{aligned}
& {{\log }_{8}}a+{{\log }_{4}}{{b}^{2}}=5 \\
& {{\log }_{4}}{{a}^{2}}+{{\log }_{8}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{3}{{\log }_{2}}a+{{\log }_{2}}b=5 \\
& {{\log }_{2}}a+\dfrac{1}{3}{{\log }_{2}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{2}}a=6 \\
& {{\log }_{2}}b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a={{2}^{6}} \\
& b={{2}^{3}} \\
\end{aligned} \right.$.
Suy ra: $ab={{2}^{6}}{{.2}^{3}}={{2}^{9}}$.
& {{\log }_{8}}a+{{\log }_{4}}{{b}^{2}}=5 \\
& {{\log }_{4}}{{a}^{2}}+{{\log }_{8}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{3}{{\log }_{2}}a+{{\log }_{2}}b=5 \\
& {{\log }_{2}}a+\dfrac{1}{3}{{\log }_{2}}b=7 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{2}}a=6 \\
& {{\log }_{2}}b=3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a={{2}^{6}} \\
& b={{2}^{3}} \\
\end{aligned} \right.$.
Suy ra: $ab={{2}^{6}}{{.2}^{3}}={{2}^{9}}$.
Đáp án A.